What is the integral from 0 to pi of sin(t)cos(nt) ?

The integral from 0 to pi of sin(t)cos(nt) is -((-1)^{-n}+1)/((n+1)(n-1))

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integral from 0 to pi of sin(t)cos(nt)

\int_{0}^{π} sin (t) cos (n t) d t

- \frac{( - 1 ) ^{- n} + 1}{( n + 1 ) ( n - 1 )}

Solution steps

\int_{0}^{π} sin (t) cos (n t) d t

Use the following identity:

cos (t) sin (s) = \frac{sin ( s + t ) + sin ( s - t )}{2}

= \int_{0}^{π} \frac{sin ( t + n t ) + sin ( t - n t )}{2} d t

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{0}^{π} sin (t + n t) + sin (t - n t) d t

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int_{0}^{π} sin (t + n t) d t + \int_{0}^{π} sin (t - n t) d t)

\int_{0}^{π} sin (t + n t) d t = \frac{( - 1 ) ^{n} + 1}{1 + n}

\int_{0}^{π} sin (t - n t) d t = \frac{( - 1 ) ^{- n} + 1}{1 - n}

= \frac{1}{2} (\frac{( - 1 ) ^{n} + 1}{1 + n} + \frac{( - 1 ) ^{- n} + 1}{1 - n})

Simplify

\frac{1}{2} (\frac{( - 1 ) ^{n} + 1}{1 + n} + \frac{( - 1 ) ^{- n} + 1}{1 - n}) : - \frac{( - 1 ) ^{- n} + 1}{( n + 1 ) ( n - 1 )}

= - \frac{( - 1 ) ^{- n} + 1}{( n + 1 ) ( n - 1 )}

\int_{0}^{1} x^{2} (2 + 2 x^{3})^{3} d x

\int_{- 3}^{2} - x^{2} - x + 6 d x

\int_{- 3}^{3} - x^{2} + 9 d x

\int_{a}^{b} \frac{1}{( x - a ) ^{n}} d x

\int_{- π}^{π} sin^{6} (x) d x

What is the integral from 0 to pi of sin(t)cos(nt) ?
The integral from 0 to pi of sin(t)cos(nt) is -((-1)^{-n}+1)/((n+1)(n-1))